Problem: The equation of a circle $C$ is $x^2+y^2-12x+12y+56 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Answer: To find the equation in standard form, complete the square. $(x^2-12x) + (y^2+12y) = -56$ $(x^2-12x+36) + (y^2+12y+36) = -56 + 36 + 36$ $(x-6)^{2} + (y+6)^{2} = 16 = 4^2$ Thus, $(h, k) = (6, -6)$ and $r = 4$.